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 Post subject: AlgebraPosted: Sun Apr 26, 2009 12:30 pm

Joined: Tue Apr 07, 2009 11:00 am
Posts: 1
How do you solve question 20 in the algebra workbook (algebra drill) page 25?
Solve for x, y, and z.

Also further explaination on questions 22 and 23.

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 Post subject: Re: AlgebraPosted: Sun Apr 26, 2009 4:47 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
Look at the first two equations:
x + y - z = 11
2x + 2y - z = 19
If we multiply the first one by 2, we can subtract it the second one and have a value for z.

2x + 2y - 2z = 22
-(2x + 2y - z = 19)
-----------------
-z = 3
z = -3

Now plug this value into the first and third equations:
x + y + 3 = 11
x - 5y + 9 = -25
Subtract the second one of these from the first:
6y - 6 = 36
6y = 42
y = 7

Then plug this back into the first equation along with the value we found for z:
x + 7 + 3 = 11
x = 1

-------------

#22.
5 * 10^(3-7)
5 * 10^(-4)
5 / 10000
.0005

#23
(x^2 - 25) / (2x + 10)
Factor the numerator and it becomes (x+5)(x-5)
Then factor a 2 out of the denominator and it becomes 2(x+5)
Now, we have (x+5) on both top and bottom, so it will cancel, leaving us with
(x-5)/2

Valerie
Veritas Help

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